Q53 / 90 · JEE Main 2024 (09 Apr Shift 1) · Mathematics
Let $\lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right.$ $\left.+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right)$ be $\frac{\pi}{k}$, using only the principal values of the inverse trigonometric functions. Then $\mathrm{k}^2$ is equal to ________
Correct answer: 32
Solution
$\begin{aligned} & \sum_{\mathrm{r}=1}^{\infty} \frac{\mathrm{n}}{\sqrt{\mathrm{n}^4+\mathrm{r}^4}}-\frac{2 \mathrm{nr}^2}{\left(\mathrm{n}^2+\mathrm{r}^2\right) \sqrt{\mathrm{n}^4+\mathrm{r}^4}} \\ & \sum_{\mathrm{r}=1}^{\infty} \frac{\frac{1}{\mathrm{n}}}{\sqrt{1+\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^4}}-\frac{2\left(\frac{1}{\mathrm{n}}\right)\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^2}{\left(1+\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^2\right) \sqrt{1+\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^4}} \\ & \Rightarrow \int_0^1 \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^4}}-\frac{2 \mathrm{x}^2 \mathrm{dx}}{\left(1+\mathrm{x}^2\right) \sqrt{1+\mathrm{x}^4}} \\ & \Rightarrow \int_0^1 \frac{1-\mathrm{x}^2}{\left(1+\mathrm{x}^2\right) \sqrt{1+\mathrm{x}^4}} d x \\ & \Rightarrow \int_0^1 \frac{\frac{1}{\mathrm{x}^2}-1}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right) \sqrt{\mathrm{x}^2+\frac{1}{\mathrm{x}^2}}} \mathrm{dx}\end{aligned}$
$\begin{aligned} & \Rightarrow-\int_0^1 \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-2}} d x \\ & x+\frac{1}{x}=t \Rightarrow 1-\frac{1}{x^2} d x=d t\end{aligned}$
$\begin{aligned} & \Rightarrow-\int_{\infty}^2 \frac{\mathrm{dt}}{\mathrm{t} \sqrt{\mathrm{t}^2-2}} \\ & \Rightarrow-\int_{\infty}^2 \frac{\mathrm{tdt}}{\mathrm{t}^2 \sqrt{\mathrm{t}^2-2}} \\ & \text { take } \mathrm{t}^2-2=\alpha^2 \\ & \mathrm{tdt}=\alpha \mathrm{d} \alpha \\ & \Rightarrow-\int_{\infty}^{\sqrt{2}} \frac{\alpha \mathrm{d} \alpha}{\left(\alpha^2+2\right) \alpha} \\ & \Rightarrow-\int_{\infty}^{\sqrt{2}} \frac{\mathrm{d} \alpha}{\alpha^2+2}\end{aligned}$
$\begin{aligned}
& \left.\Rightarrow \frac{-1}{\sqrt{2}} \tan ^{-1} \frac{\alpha}{\sqrt{2}}\right]_{\infty}^{\sqrt{2}} \\
& \Rightarrow \frac{-1}{\sqrt{2}}\left\{\tan ^{-1} 1\right\}+\frac{1}{\sqrt{2}} \tan ^{-1} \infty \\
& \Rightarrow \frac{1}{\sqrt{2}}\left\{\frac{\pi}{2}-\frac{\pi}{4}\right\} \\
& \Rightarrow \frac{\pi}{4 \sqrt{2}}=\frac{\pi}{\mathrm{K}}
\end{aligned}$
So $\mathrm{K}=4 \sqrt{2}$ $\mathrm{K}^2=32$
So $\mathrm{K}=4 \sqrt{2}$ $\mathrm{K}^2=32$
Hence, the correct answer is 32.